What is a good lower-bound for $P(t|a^Tx| \le \|x\|^2)$, where $a$ is a fixed unit vector in $\mathbb R^m$, $x \sim N(0, I_m)$, and $t > 0$?

Question

Let $a$ be a fixed unit vector in $\mathbb R^m$ and $x$ be a random vector in $\mathbb R^m$ with iid entries from $N(0, 1)$. Let $t > 0$.

Question. What is a good lower-bound for $P(t|a^Tx| \le \|x\|^2)$ ?

A crude bound

One computes $P(t|a^Tx| \le \|x\|^2) \ge P(|a^Tx| \le t,\,\|x\| > t) \ge 1 - P(|a^Tx| > t) - P(\|x\|^2 \le t^2)$, where the last inequality is a union bound. On the other hand, using this anti-concentraiton result, one can show that

$$ P(\|x\|^2 \le t^2) = P(\|x\|^2 \le (t^2/m)m) \le \frac{Ct}{\sqrt{m}}, \forall t \in (0,\sqrt{m}/C), $$

where $C > 0$ is an absolute constant ($C = \sqrt{e} \approx 1.65$ works). Also, by standard Gaussian concentraiton arguments, we have $P(|a^Tx| > t) \le 2e^{-t^2/2}$. Thus if $0 < t \le \sqrt{m}/C$, we have

$$ P(t|a^Tx| \le \|x\|^2) \ge 1 - 2e^{-t^2/2} - \frac{Ct}{\sqrt{m}}. $$

Answer 1

It turns out we can get a decent bound using standard concentration results for polynomials.

[Carbery and Wright (2001)] Let $f$ be a $m$-variate polynomial of degree $d$. Then $$ \sup_{u \in \mathbb R} P(|f(X) - u| \le \varepsilon) \le \mathcal O(d)\varepsilon^{1/d}. $$ In particular, if $P$ is a nonzero psd matrix with, then $$ P(X^TPX \le \varepsilon) \le \sqrt{\frac{2\varepsilon}{\mbox{trace}(P)}}. $$

Now, for any $s > 0$, we have $t|\langle a,X\rangle| \le st^2\langle a, X\rangle^2 + 1/s$ and so

$$ P(\|X\|^2 \ge |\langle a, X\rangle|) \ge \sup_{t > 0}P(\|X\|^2 \ge st^2\langle a,X\rangle ^2 + 1/s). $$

Let $f_s(X) := \|X\|^2 - st^2\langle a,X\rangle^2$. A simple computation reveals that $f_S(X) = X^TX - st^2X^Taa^TX = X^TP_sX$, where $P_s := I-st^2 aa^T$, which is symmetric. Moreover, if $s \le t^{-2}$, then $P_s$ is psd with trace $m - st^2$. Therefore, using the CW inequality with $f(X) = X^TP_sX \ge 0$, $d=2$, $\varepsilon = 1/s$, and $u=0$ gives

$$ \begin{split} P(\|X\|^2 \ge t|\langle a, X\rangle|) &\ge \sup_{0 < s \le t^{-2}}P(\|X\|^2 \ge s\langle a,X\rangle ^2 + 1/s) = \sup_{0 < s \le t^{-2}}P(X^TP_sX \ge 1/s)\\ &\ge \sup_{0 < s \le t^{-2}}1 - C\frac{1}{\sqrt{s}\sqrt{m-st^2}} = 1 - C\frac{t}{\sqrt{m-1}}, \end{split} $$

where $C > 0$ is an absolute constant ($C = \sqrt{2}$ works!).